Number Theory: An Introduction to Mathematics

516 XII Elliptic Functions

where

α′={γ′+(γ′^2 +δ′^2 )^1 /^2 }^1 /^2 /

√

2 , 2 α′β′=δ′.

Thusα′andβ′are positive for 0<y<1, and

K( 1 −λ)=A′−iB′,

whereA′>0,B′>0.
We will now show that the period ratioiK( 1 −λ)/K(λ)is not real by showing that
the quotientK( 1 −λ)/K(λ)has positive real part. Since this is equivalent to showing
that

AA′−BB′> 0 ,

it is sufficient to show thatαα′−ββ′>0forallx,y∈( 0 , 1 ). The inequality is cer-
tainly satisfied for allx,ynear 0, sinceα→1,β→0asx→0andα′→1,β′→ 0
asy→0. Thus we need only show that we never haveαα′=ββ′.But

2 α^2 =(γ^2 +δ^2 )^1 /^2 +γ, 2 β^2 =(γ^2 +δ^2 )^1 /^2 −γ,

with analogous expressions for 2α′^2 , 2 β′^2. Hence, ifαα′=ββ′, then by squaring we
obtain

[(γ^2 +δ^2 )^1 /^2 +γ][(γ′^2 +δ′^2 )^1 /^2 +γ′]=[(γ^2 +δ^2 )^1 /^2 −γ][(γ′^2 +δ′^2 )^1 /^2 −γ′],

which reduces to

γ(γ′^2 +δ′^2 )^1 /^2 =−γ′(γ^2 +δ^2 )^1 /^2.

Squaring again, we obtainγ^2 δ′^2 =γ′^2 δ^2. Since the previous equation shows thatγ
andγ′do not have the same sign, it follows that

γδ′+γ′δ= 0.

Givingγ,δ,γ′,δ′their explicit expressions, this takes the formv(x+y−xy)=0.
Hencex( 1 −y)+y=0, which is impossible if 0<y<1andx>0.
The relationS[u(z)]=z,whereu(z)is defined by (23), shows that the elliptic
integral of the first kind isinvertedby the elliptic functionS(u).Wemayusethisto
simplify other elliptic integrals. The change of variablesx=S(u)replaces the integral

∫

R(x)dx/gλ(x)^1 /^2

by

∫

R[S(u)]du. Following Jacobi, we take

E(u):=

∫u

0

[1−λS(v)]dv (24)